I have one chance over $1,000,000$ to win a lottery. If I buy the same number of ticket with different combination on each, I have $100\%$ chance of winning. What if my ticket number are randomly chosen ?
What is the probability of winning a lottery (that has $1,000,000$ possible combination) if I choose randomly the $1,000,000$ ticket's number ?
Thank you
On
I'll assume that the ticket numbers are chosen independently and at random, and each has a $1/1000000$ chance of winning.
In this case, each ticket has a $999999/1000000$ chance of losing. If you buy $1000000$ tickets, the probability that all of them lose is $$\left(\frac{999999}{1000000}\right)^{1000000} \, \approx \, 0.367879$$
The probability that at least one of them wins is equal to $1$ minus the probability that all of them lose:
$$1 - \left(\frac{999999}{1000000}\right)^{1000000} \,\approx \,0.63212 \,\approx \,\boxed{\,63.2\%\,}\,$$
As the answer of Zubin Mukerjee has already pointed out the probability is
$$1-\left(\frac{999999}{1000000}\right)^{1000000} $$
You can use the relation $\lim\limits_{n\to \infty }(1-\frac{x}{n})^n=e^{-x}$ to get an appropiate approximation.
$\frac{999999}{1000000}=1-\frac{1}{1000000}$. Therefore we can write
$1-\left(\frac{999999}{1000000}\right)^{1000000}=1-\left( 1-\frac{1}{1000000}\right)^{1000000}\approx \boxed{1-e^{-1}}=63.21\%$